Question

Tension in rod of length $L$ and mass $M$ at a distance $y$ from $F_1$ when the rod is acted on by two unequal forces $F_1$ and $F_2$ where ($F_2<F_1$) at its ends is :-

• A. $F_1(1-y/L)+F_2\left(y/L\right)$
• B. $F_2(1-y/L)+F_1\left(y/L\right)$
• C. $F_1(1+y/L)+F_2\left(y/L\right)$
• D. $F_2(1+y/L)+F_1\left(y/L\right)$

Answer 1

The correct option is A $F_1(1-y/L)+F_2\left(y/L\right)$

Let $m$ be the mass per unit length.
As $F_2<F_1$, so the acceleration of the system in left direction (shown in the figure).
For part AP, $mya=F_1-T....\left(1\right)$
for part PB, $m(L-y)a=T-F_2....\left(2\right)$
$\left(2\right)/\left(1\right),(\frac{L}{y}-1)=\frac{T-F_2}{F_1-T}$
or, $(\frac{L}{y}-1)F_1-(\frac{L}{y}-1)T=T-F_2$
or, $T(\frac{L}{y}-1+1)=(\frac{L}{y}-1)F_1+F_2$
or, $T=(1-\frac{y}{L})F_1+F_2\left(\frac{y}{L}\right)$